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        <p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/distinct-subsequences-ii/">原题连接</a></p>
<h1 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h1><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">给定一个字符串，计算这个字符串不同非空子序列的个数，对结果取余1000000007</span><br></pre></td></tr></table></figure>
<span id="more"></span>
<p><img src="image-20211019225338113.png" alt="image-20211019225338113"></p>
<h1 id="分析"><a href="#分析" class="headerlink" title="分析"></a>分析</h1><ol>
<li>当然可以dfs子集，用set去重，但是数据范围限制字符串最长为2000，不能接受指数的复杂度。</li>
<li>考虑动态规划</li>
</ol>
<h2 id="动态规划思路"><a href="#动态规划思路" class="headerlink" title="动态规划思路"></a>动态规划思路</h2><ol>
<li>针对每一个字符str(i)，都有选或者不选两种可能</li>
</ol>
<p><img src="image-20211019230128020.png" alt="image-20211019230128020"><br>所以转移方程为</p>
<script type="math/tex; mode=display">
f(i)=2f(i-1)-num</script><ol>
<li>重点考虑num的值。</li>
<li>如果字符串不重复，则num=0，f(i)=2f(i-1)</li>
<li>如果存在重复</li>
</ol>
<p>设当前字符为 @，且集合 i-1 中必定包含字符@。<br><figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">假设集合i-1中包含 **@，***@,则其一定包含**和***，那么和当前组合一定会产生**@和***@这两个重复项。</span><br></pre></td></tr></table></figure><br>因此，</p>
<script type="math/tex; mode=display">
num=f[last[c]-1]</script><p>last(c)为字符c最近出现的位置。</p>
<p><img src="image-20211019230824868.png" alt="image-20211019230824868"></p>
<h1 id="代码实现"><a href="#代码实现" class="headerlink" title="代码实现"></a>代码实现</h1><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">distinctSubseqII</span><span class="params">(String s)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// f(i)，以s[i]为结尾的所有子序列</span></span><br><span class="line">        <span class="keyword">char</span> []str= s.toCharArray();</span><br><span class="line">        <span class="keyword">int</span> []f=<span class="keyword">new</span> <span class="keyword">int</span>[str.length+<span class="number">1</span>];</span><br><span class="line">        <span class="keyword">int</span> []last=<span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">26</span>];</span><br><span class="line">        Arrays.fill(last,-<span class="number">1</span>);</span><br><span class="line">        f[<span class="number">0</span>]=<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">int</span> mod=<span class="number">1000000007</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i&lt;f.length;i++)&#123;</span><br><span class="line">            <span class="keyword">int</span> x=str[i-<span class="number">1</span>]-<span class="string">&#x27;a&#x27;</span>;</span><br><span class="line">            <span class="keyword">if</span>(last[x]==-<span class="number">1</span>)&#123;</span><br><span class="line">                f[i]=(f[i-<span class="number">1</span>]*<span class="number">2</span>)%mod;</span><br><span class="line">            &#125;<span class="keyword">else</span>&#123;   </span><br><span class="line">                f[i]=f[i-<span class="number">1</span>]*<span class="number">2</span>%mod-f[last[x]-<span class="number">1</span>];</span><br><span class="line">            &#125;</span><br><span class="line">            f[i]=f[i]%mod;</span><br><span class="line">            last[x]=i;</span><br><span class="line">            <span class="comment">// System.out.println(Arrays.toString(f));</span></span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> (f[str.length]-<span class="number">1</span>+mod)%mod;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
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